Q235 / 270 · NDA 2016 (Phase 2) · Mathematics

Paragraph: Let $f(x)=\left\{\begin{array}{ll}\frac{e^x-1}{x}, & x \gt 0 \\ 0, & x=0\end{array}\right.$ be a real valued function.
Question: Which of the following statements is/are correct? 1. $f(x)$ is right continuous at $x=0$. 2. $f(x)$ is discontinuous at $x=1$. Select the correct answer using the code given below:

  • A
    1 only
  • Correct answer
    B
    2 only
  • C
    Both 1 and 2
  • D
    Neither 1 nor 2

Solution

For right hand continuity at $x=0$ $\begin{aligned} & \begin{aligned} \mathrm{RHL} & =\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ & =\lim _{h \rightarrow 0} \frac{e^h-1}{h}=\lim _{h \rightarrow 0} \frac{\left(1+h+\frac{h^2}{2!}+\ldots\right)-1}{h} \\ & =\lim _{h \rightarrow 0} \frac{h+\frac{h^2}{2!}+\frac{h^3}{3!}+\ldots}{h} \\ & =\lim _{h \rightarrow 0} 1+\frac{h}{2!}+\frac{h^2}{3!}+\ldots . .=1 \end{aligned} \\ & \begin{array}{l} f(0)=0 \\ \Rightarrow f(x) \text { is not right continuous at } x=0 . \\ \text { For discontinuity at } x=1 \end{array} \\ & \text { RHL }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \end{aligned}$ $\begin{aligned} & =\lim _{h \rightarrow 0} \frac{e^{1+h}-1}{1+h} \\ & =\lim _{h \rightarrow 0} \frac{\left(1+(1+h)+\frac{(1+h)^2}{2}+\ldots .\right)-1}{1+h} \\ & =\lim _{h \rightarrow 0} \frac{(1+h)+\frac{(1+h)^2}{2!}+\frac{(1+h)^3}{3!}+\ldots .}{(1+h)} \\ & =\lim _{h \rightarrow 0} 1+\frac{(1+h)}{2!}+\frac{(1+h)^2}{3!}+\ldots . \\ & =1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots . \\ & \mathrm{LHL}=\lim _{x \rightarrow \mathrm{I}^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ & =\lim _{h \rightarrow 0} \frac{e^{(1-h)}-1}{1-h} \\ & =\lim _{h \rightarrow 0} \frac{\left(1+(1-h)+\frac{(1-h)^2}{2!}+\ldots .\right)-1}{(1-h)} \\ & =\lim _{h \rightarrow 0} 1+\frac{(1-h)}{2!}+\frac{(1-h)^2}{3!}+\ldots . \\ & =1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots . \\ & f(1)=\frac{e^1-1}{1}=\left(1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots .\right)-1 \end{aligned}$ $=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots$ $\Rightarrow \mathrm{RHL} \neq f(1), \mathrm{LHL} \neq f(1)$ So $f$ is discontinuous.

Hence, the correct answer is option: 'B'.

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